Question

An insulated mixing chamber receives 0.5 kg/s of steam at 3 MPa and 300°C through one inlet, and saturated liquid water at 3 MPa through the other inlet. The combined stream exits with a quality of 80% and a pressure of 3 MPa. A fan, using 10 kW of power, is used to aid the mixing process. Determine the mass flow rate of the saturated liquid stream entering the mixing chamber

Answer 1

`\dot m_(2) = 0.199\,(kg)/(s)`

Step-by-step explanation:

The mixing chamber can be modelled by applying the First Law of Thermodynamics:

`\dot W_(in)+\dot m_(1)\cdot h_(1) +\dot m_(2) \cdot h_(2) - \dot m_(3)\cdot h_(3) = 0`

Since that mass flow rate of water at inlet 1 is the only known variable, the expression has to be simplified like this:

`(\dot W_(in))/(\dot m_(1)) + h_(1)+ y\cdot h_(2) - z\cdot h_(3) = 0`

Besides, the following expression derived from the Principle of Mass Conservation is presented below:

`1 + y = z`

Then, the expression is simplified afterwards:

`(\dot W_(in))/(\dot m_(1)) + h_(1)+ y\cdot h_(2) - (1+y)\cdot h_(3) = 0`

`(\dot W_(in))/(\dot m_(1)) +h_(1) - h_(3) + y\cdot (h_(2)-h_(3)) = 0`

Specific enthalpies are obtained from steam tables and described as follows:

State 1 (Superheated vapor)

`h = 2994.3\,(kJ)/(kg)`

State 2 (Saturated liquid)

`h = 1008.3\,(kJ)/(kg)`

State 3 (Liquid-Vapor mixture)

`h = 2444.22\,(kJ)/(kg)`

The ratio of the stream at state 2 to the stream at state 1 is:

`y = ((\dot W_(in))/(\dot m_(1))+h_(1)-h_(3))/(h_(3)-h_(2))`

`y = ((10\,kW)/(0.5\,(kg)/(s) )+2994.3\,(kJ)/(kg)-2444.22\,(kJ)/(kg) )/(2444.22\,(kJ)/(kg)-1008.3\,(kJ)/(kg) )`

`y = 0.397`

The mass flow rate of the saturated liquid is:

`\dot m_(2) = y\cdot \dot m_(1)`

`\dot m_(2) = 0.397\cdot (0.5\,(kg)/(s) )`

`\dot m_(2) = 0.199\,(kg)/(s)`